What is the integral from 0 to 9 of sqrt(9+3x) ?

The integral from 0 to 9 of sqrt(9+3x) is 42

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integral from 0 to 9 of sqrt(9+3x)

\int_{0}^{9} 9 + 3 x d x

42

Solution steps

\int_{0}^{9} 9 + 3 x d x

Apply u-substitution

= \int_{9}^{36} \frac{u}{3} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int_{9}^{36} u d u

Apply the Power Rule

= \frac{1}{3} [\frac{2}{3} u^{\frac{3}{2}}]_{9}^{36}

Compute the boundaries:

126

= \frac{1}{3} \cdot 126

Simplify

= 42

What is the integral from 0 to 9 of sqrt(9+3x) ?
The integral from 0 to 9 of sqrt(9+3x) is 42